Find Volume of bound solid? (around the x-axis)?

Question

y=√x+ (1)

y=x+1

Answer 1

What are the limits in x?

I assume we start at x = 0.

The function will be 0 and 2 when x = 1 and after that it will be on both sides of the x-axis if we consider both positive and negative roots.

So it appears that you want the volume between x = 0 and x =1.

If I only use the positive root then the bounds of the volume are:

Upper bound is SQRT(x) + 1 = y1

Lower bound is -SQRT(x) + 1 = y2

And rotated about the x-axis so:

dV = [2*pi*(y1 - y2)]dx

dV = [2*pi*(2SQRT(x)]dx

V = 4*pi[x^(3/2)]/(3/2)

V = (8*pi/3)x^(3/2) ................... evaluate between 0 and 1

V = (8*pi/3)[(1) - (0)]

V = 8*pi/3

Check if answer is reasonable.

This will be approximately the volume of a truncated cone with radius 2 and height 1 Overall height 3) minus the volume of a cone of radius 1 and height 1.

V1 = (1/3)pi(2^2)(3) - (1/3)pi(1.3)^2(2)

V1 = (1/3)pi[12 - 3.38] = 8.62pi/3

V2 = (1/3)pi(1)(1) = (1/3)pi

V1 - V2 = 7.62pi/3

So I think my answer looks good